
\prob{0099}{根式方程V}

求方程
\[ \sqrt{7x^2 + 9x + 13} + \sqrt{7x^2 - 5x + 13} = 7x \]
的实数根。
\problabels{yellow/代数, green/方程相关问题}

\textit{JYH提供的题目。}

\ans{$x = 12/7$}

\subsection{换元}

令
\[ p = \sqrt{7x^2 + 9x + 13}, q = \sqrt{7x^2 - 5x + 13} \]
则有
\[ p + q = 7x = \frac12\left(p^2 - q^2\right) = \frac12(p + q)(p - q) \]
显然$p + q \ne 0$，故$p - q = 2, p + q = 7x$，于是知
\[ 2p = 7x + 2 \Rightarrow 4\left(7x^2 + 9x + 13\right) = (7x + 2)^2 \]
解得$x_1 = 12/7, x_2 = -4/3$。

经检验，$x_1 = 12/7$为原方程的解，$x_2 = -4/3$为增根。故$x = 12/7$。
